Question

5. In Problems 1-30 use appropriate algebra and Theorem 7.2.1 to find the given inverse Laplace transform.

${\mathcal{L}}^{\mathbf{-}\mathbf{1}}\left\{\frac{{(s+1)}^3}{s^4}\right\}$

Short answer

$1+3\mathrm{t}+\frac{3}{2}{\mathrm{t}}^2+\frac{1}{6}{\mathrm{t}}^3$

Step-by-step solution

Define the formulas for inverse Transform

Consider the formula for inverse Laplace given below:

1. $1={\mathcal{L}}^{-1}\left\{\frac{1}{\mathrm{s}}\right\}$
2. ${\mathrm{t}}^{\mathrm{n}}={\mathcal{L}}^{-1}\left\{\frac{\mathrm{n}!}{{\mathrm{s}}^{\mathrm{n}+1}}\right\},\mathrm{n}=1,2,3,\dots$
3. ${\mathrm{e}}^{\mathrm{at}}={\mathcal{L}}^{-1}\left\{\frac{1}{\mathrm{s}-\mathrm{a}}\right\}$
4. $\mathrm{sinkt}={\mathcal{L}}^{-1}\left\{\frac{\mathrm{k}}{{\mathrm{s}}^2+{\mathrm{k}}^2}\right\}$
5. $\mathrm{coskt}={\mathcal{L}}^{-1}\left\{\frac{\mathrm{s}}{{\mathrm{s}}^2+{\mathrm{k}}^2}\right\}$
6. $\mathrm{sinhkt}={\mathcal{L}}^{-1}\left\{\frac{\mathrm{k}}{{\mathrm{s}}^2-{\mathrm{k}}^2}\right\}$
   
7. $\mathrm{coshkt}={\mathcal{L}}^{-1}\left\{\frac{\mathrm{s}}{{\mathrm{s}}^2-{\mathrm{k}}^2}\right\}$
   

Determine the inverse Laplace transform:

Resolve the equation as:

$$\begin{gathered} L^{-1}\left\{\frac{{(s+1)}^3}{s^4}\right\}=L^{-1}\left\{\frac{s^3}{s^4}+\frac{3s^2}{s^4}+\frac{3s}{s^4}+\frac{1}{s^4}\right\} \\ =L^{-1}\left\{\frac{1}{s}\right\}+L^{-1}\left\{\frac{3}{s^2}\right\}+L^{-1}\left\{\frac{3}{s^3}\right\}+L^{-1}\left\{\frac{1}{s^4}\right\} \end{gathered}$$

From the inverse Laplace transform rules and formulas:

$$\begin{gathered} L^{-1}\left\{\frac{{(s+1)}^3}{s^4}\right\}=1+3t+3\frac{1}{2!}{t}^2+\frac{1}{3!}{t}^3 \\ =1+3t+\frac{3}{2}{t}^2+\frac{1}{6}{t}^3 \end{gathered}$$